Question

How do you integrate (x+3)ln^2(x+3)dx?

Answer 1

`I=(x+3)^2/4[2ln^2(x+3)-2ln(x+3)+1]+c`

Explanation:

Here,

`I=int(x+3)ln^2(x+3)dx`

`I=int(x+3)[ln(x+3)]^2dx`

We take,

`ln(x+3)=t=>x+3=e^t=>dx=e^tdt`

So,

`I=int(e^t)[t]^2e^tdt`

`:.I=intt^2*e^(2t)dt`

`"Using "color(blue)"Integration by parts : "`

`I=t^2inte^(2t)dt-int(d/(dt)(t^2) xx inte^(2t)dt)dt`

`=t^2*(e^(2t))/2-int 2t*(e^(2t))/2dt`

`=t^2*(e^(2t))/2-int t*(e^(2t))dt`

Again, `"using "color(blue)"Integration by parts : "`in second integral,

`I=t^2*(e^(2t))/2-[t*(e^(2t))/2-int(1)(e^(2t)/2)dt]`

`=t^2*(e^(2t))/2-t*e^(2t)/2+1/2inte^(2t)dt`

`=t^2*(e^(2t))/2-t*e^(2t)/2+1/2*e^(2t)/2+c`

`:.I=e^(2t)/4[2t^2-2t+1]+c`

`=>I=(x+3)^2/4[2ln^2(x+3)-2ln(x+3)+1]+c`

Note:

`ln(x+3)=t`

`=>[ln(x+3)]^2=t^2`

`=>ln^2(x+3)=t^2`

Also,

`(i)ln^2(x+3)=[ln(x+3)]^2=ln(x+3)xxln(x+3),and`

`(ii)ln(x+3)^2=2ln(x+3)`

`=>ln^2(x+3)!=ln(x+3)^2`